Engineering, 22.04.2020 23:20 blumkemax8

An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used

Answers: 1

Show answers

Answers

Answer from: GERIANNCAYER

Complete Question

The diagram for this question is shown on the first uploaded image

The largest offset that can be used is $h = 0.455 \ in$

Explanation:

From the question we are told that

The diameter of the metal tube is $d_m = 0.75 \ in$

The thickness of the wall is $D = 0.08 \ in$

Generally the inner diameter is mathematically evaluated as

d_i = d_m -2D

= 0.75 - 2(0.08)

$= 0.59 \ in$

Generally the tube's cross-sectional area can be evaluated as

$a = \frac{\pi}{4} (d_m^2 - d_i^2)$

$= \frac{\pi}{4} (0.75^2 - 0.59^2)$

$= 0.1684 \ in^2$

Generally the maximum stress of the metal is mathematically evaluated as

$\sigma = \frac{P}{A}$

$\sigma = \frac{P}{ 0.1684}$

The diagram showing when the stress is been applied is shown on the second uploaded image

Since the internal forces in the cross section are the same with the force P and the bending couple M then

M = P * h

Where h is the offset

The maximum stress becomes

$\sigma_n = \frac{P}{A} + \frac{M r_m }{I}$

Where r_m is the radius of the outer diameter which is evaluated as

$r_m = \frac{0.75}{2}$

$r_m = 0.375 \ in$

and I is the moment of inertia which is evaluated as

$I = \frac{\pi}{64} (d_m^4 - d_i^4 )$

$= \frac{\pi}{64}(0.75^4 - 0.59^4)$

$= 0.009583 \ in^4$

So the maximum stress becomes

$\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}$

Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress

$\sigma ' = 4 \sigma$

=> $\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ]$

The P would cancel out

$\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}$

5.94 + 39.13h = 23.753

39.13h = 17. 813

$h = 0.455 \ in$

An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickn

Answer from: Quest

answer:

i lost in getting someone back.

i learned that people are really two-faced and no matter how angelic someone appears they always have a devil side and that you cant trust anyone anymore no matter how dear they are to you.

if your wondering who it waz it waz mah still miss her tho

~batmans wife dun dun

Answer from: Quest

answer: hi

explanation:

Another question on Engineering

Engineering, 04.07.2019 18:10

Determine whether or not it is possible to compress air adiabatically from k to 140 kpa and 400 k. what is the entropy change during this process?

Answers: 3

Answer

Engineering, 04.07.2019 18:10

Burgers vector is generally parallel to the dislocation line. a)-true b)-false

Answers: 2

Answer

Engineering, 04.07.2019 18:10

The higher the astm grain size number, the finer the gran is. a)-true b)-false

Answers: 2

Answer

Engineering, 04.07.2019 18:10

Awall of 0.5m thickness is to be constructed from a material which has average thermal conductivity of 1.4 w/mk. the wall is to be insulated with a material having an average thermal conductivity of 0.35 w/mk so that heat loss per square meter shall not exceed 1450 w. assume inner wall surface temperature of 1200°c and outer surface temperature of the insulation to be 15°c. calculate the thickness of insulation required.

Answers: 3

Answer

You know the right answer?

An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickn...

Questions